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# 03hw - Solutions to problem set 3(141A Sp07 1 1(ISSP 3.3...

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Solutions to problem set 3 (141A Sp07) 1 1. (ISSP 3.3) Basis of two unalike atoms. At k = π/ a , equation (20) becomes 2 - ω 2 M 1 C ! u = 0 2 - ω 2 M 2 C ! v = 0 Note that the u and v sub-lattices are decoupled at this value of k ; u can change without changing v , and vice versa. There are two solutions to the above equations: u = 0, ω 2 = 2 C / M 2 and v arbitrary; and v = 0, ω 2 = 2 C / M 1 and u arbitrary. 2. (ISSP 3.4) Kohn anomaly. Equation (16a) gives ω 2 = 2 A M X p = 1 sin pk 0 a pa (1 - cos( pka )) = 2 A Ma X p = 1 sin pk 0 a p - 1 2 sin p ( k 0 + k ) a p - 1 2 sin p ( k 0 - k ) a p ! Gradshteyn and Ryzhik ( Table of integrals, series and products )1.441 gives X p = 1 sin α p p = π - α 2 0 < α < 2 π (you can prove this by writing the sin as the sum of two exponentials, and then using the taylor expansion of log (1 + x ).) So for k 0 > k , we have ω 2 = 2 A Ma X p = 1 sin pk 0 a p - 1 2 sin p ( k 0 + k ) a p - 1 2 sin p ( k 0 - k ) a p ! = 2 A Ma π - k 0 a 2 - 1 2 π - ( k 0 + k ) a 2 - 1 2 π - ( k 0 - k ) a 2 ! = 0 Now consider the case where k 0 < k . In order to apply the formula from the tables we need the argument of each sin function to be positive. Then ω 2 = 2 A Ma X p = 1 sin pk 0 a p - 1 2 sin p ( k 0 + k ) a p - 1 2 sin p ( k 0 - k ) a p !

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03hw - Solutions to problem set 3(141A Sp07 1 1(ISSP 3.3...

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